Question

A uniform magnetic field \(\overrightarrow{B}=B_0\hat{z}\), where \(B_0\gt0\) exists as shown in the figure. A charged particle of mass m and charge \(q(q\gt0)\) is released at the origin, in the yz-plane, with a velocity \(\overrightarrow v\) directed at an angle \(\theta\)= 45° with respect to the positive z-axis. Ignoring gravity, which one of the following is TRUE.

• A. The initial acceleration \(\overrightarrow a=\frac{qvB_0}{\sqrt 2m}\hat x\)
• B. The initial acceleration \(\overrightarrow a=\frac{qvB_0}{\sqrt 2m}\hat y\)
• C. The particle moves in a circular path
• D. The particle continues in a straight line with constant speed

Answer 1

The Correct Option is A

To solve this problem, we need to analyze the motion of a charged particle in a magnetic field. Let's break down the given setup:

• The magnetic field is \(\overrightarrow{B}=B_0\hat{z}\).
• The charge \(q\) is positive, and the particle mass is \(m\).
• The particle is released with velocity \(\overrightarrow{v}\) at an angle \(\theta = 45^\circ\) with respect to the z-axis.

The velocity can be resolved into components:

• \(\overrightarrow{v}_z = v \cos \theta \, \hat{z} = v \frac{1}{\sqrt{2}} \, \hat{z}\)
• \(\overrightarrow{v}_y = v \sin \theta \, \hat{y} = v \frac{1}{\sqrt{2}} \, \hat{y}\)

Using the right-hand rule for a velocity component perpendicular to the magnetic field:

• The magnetic force is given by \(\overrightarrow{F} = q (\overrightarrow{v} \times \overrightarrow{B})\).

In this setup, the z-component of velocity does not contribute to the force as it is parallel to the magnetic field. Only the y-component of velocity contributes:

• \(\overrightarrow{F} = q \left(v \frac{1}{\sqrt{2}} \hat{y} \times B_0 \hat{z}\right) = q v \frac{1}{\sqrt{2}} B_0 (-\hat{x}) = -\frac{q v B_0}{\sqrt{2}} \hat{x}\)

The acceleration of the particle is given by:

• \(\overrightarrow{a} = \frac{\overrightarrow{F}}{m} = -\frac{q v B_0}{\sqrt{2} m} \hat{x}\)

This shows the acceleration is initially in the x-direction, matching option:

The initial acceleration \(\overrightarrow{a}=\frac{qvB_0}{\sqrt{2}m}\hat{x}\)

Conclusion: The correct answer is that the initial acceleration is \(\overrightarrow{a}=\frac{qvB_0}{\sqrt{2}m}\hat{x}\).